Mark me as Brainliest Please
Answer:
Step-by-step explanation:
From the information given,
Number of personnel sampled, n = 85
Mean or average = 6.5
Standard deviation of the sample = 1.7
We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.
For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.
z×standard deviation/√n
= 1.96 × 6.5/√85
= 1.38
The confidence interval for the mean number of years spent before promotion is
The lower end of the interval is 6.5 - 1.38 = 5.12 years
The upper end is 6.5 + 1.38 = 7.88 years
Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years
Answer:
im too un intellegenced for this
Step-by-step explanation:
The first odd number is 1. The next is 2 more than the previous one.
It's an arithmetic sequence.
The general term of an arithmetic sequence:
a_n - n-th term
a_1 - first term
d - common difference.
We have
Substitute:
<h3>Answer: 137th odd number is equal 273.</h3>
